Real-Life Examples of Quantum Mechanic

At what velocity does a proton have a 6.00-fm wavelength (about the size of a nucleus)? Assume the proton is nonrelativistic. (1 femtometer $={10}^{-15}m.$ )

Does photon behave as both a particle and a wave? Explain.

A 100 MeV photon collides with a resting prion. Calculate the maximum energy loss that the photon can suffer.

JC-23) de Broglie Waves A photon and a particle have the same wavelength.
a) How do their linear momenta compare?
b) How does the photon's energy compare to the particle's total energy?
c) How does the photon's energy compare to the particle's kinetic energy?

If an incident photon of a wavelength A has scattered with an angle e after it collides with an electron in an atom, what is the shift in its wave length?

Landauer's principle vs Wien's displacement law
Can we argue based on Landauer's principle that if one bit information is changed inside a blackbody, the total radiated energy should be at least or in order of kTln2? If it is so, can we also argue that this energy should be distributed over all the modes of the cavity? Furthermore, can it also be argued that this contradicts with the Wien's displacement law which says the total energy should be in the order of $k\frac{T^4}{h^4}$ (as integrated using Mathematica 8.0)?

Wien's Displacement and Rayleigh-Jeans Law from Planck's Law
I've derived Planck's Law for frequency from his law for wavelength, and I got this:
$u(f)=\frac{8\mathrm{\pi <!--\; \pi \; -->}{f}^2}{c^3}\frac{hf}{e^{\frac{hf}{kT}}-<!--\; -\; -->1}$
I just have a quick question about this. This problem says I need to find the low-frequency limit (which will lead to the Rayleigh-Jeans), and I need to take the high-frequency limit, which is supposed to lead to Wien's distribution.
The thing is I have no idea what the high, or low frequency limit even means, I've looked in my book, I've looked online, I'm not sure exactly what it means.

How do we know that dark matter is dark?
How do we know that dark matter is dark, in the sense that it doesn't give out any light or absorb any? It is impossible for humans to be watching every single wavelength. For example, what about wavelengths that are too big to detect on Earth?

Why is the de Broglie equation as well as the Schrodinger equation is correct for massive particle?
Starting from special relativity, here I see the de Broglie approximation is valid only if $m_0=0$
Derivation:
$E^2=P^2{C}^2+m_0^2{C}^4$. Here we put Plank-Einstein relation $E=h\mathrm{\nu <!--\; \nu \; -->}=h\frac{C}{\mathrm{\lambda <!--\; \lambda \; -->}}$. Finally,
$\mathrm{\lambda <!--\; \lambda \; -->}=\frac{h}{\sqrt{P^2+m_0^2{C}^2}}(1)$
If $m_0=0$ then $\mathrm{\lambda <!--\; \lambda \; -->}=\frac{h}{p}$ (de Broglie approximation).
Furthermore we know the Schrodinger equation was derived by assuming that the de Broglie approximation is true for all particles, even if $m_0\ne <!--\; \ne \; -->0$. But if we take special relativity very strictly then this approximation looks incorrect.
In addition, if we try to derive the Schrodinger equation from the exact relation found in '1', we find completely different equation. For checking it out, lets take a wave function-
$\mathrm{\Psi <!--\; \Psi \; -->}=A{e}^{i(\frac{2\mathrm{\pi <!--\; \pi \; -->}}{\mathrm{\lambda <!--\; \lambda \; -->}}x-<!--\; -\; -->\mathrm{\omega <!--\; \omega \; -->}t)}=A{e}^{i(\frac{\sqrt{P^2+m_0^2{C}^2}}{\mathrm{\hslash <!--\; \hslash \; -->}}x-<!--\; -\; -->\frac{E}{\mathrm{\hslash <!--\; \hslash \; -->}}t)}$ (putting $\mathrm{\lambda <!--\; \lambda \; -->}$ from '1', $\frac{h}{2\mathrm{\pi <!--\; \pi \; -->}}=\mathrm{\hslash <!--\; \hslash \; -->}$ and $E=\mathrm{\hslash <!--\; \hslash \; -->}\mathrm{\omega <!--\; \omega \; -->}$).
Then, $\frac{{\mathrm{\partial <!--\; \partial \; -->}}^2\mathrm{\Psi <!--\; \Psi \; -->}}{\mathrm{\partial <!--\; \partial \; -->}{x}^2}=-<!--\; -\; -->\frac{P^2+m_0^2{C}^2}{{\mathrm{\hslash <!--\; \hslash \; -->}}^2}\mathrm{\Psi <!--\; \Psi \; -->}=-<!--\; -\; -->\frac{E^2}{C^2{\mathrm{\hslash <!--\; \hslash \; -->}}^2}\mathrm{\Psi <!--\; \Psi \; -->}$
$⟹<!--\; ⟹\; -->E^2\mathrm{\Psi <!--\; \Psi \; -->}=-<!--\; -\; -->C^2{\mathrm{\hslash <!--\; \hslash \; -->}}^2\frac{{\mathrm{\partial <!--\; \partial \; -->}}^2\mathrm{\Psi <!--\; \Psi \; -->}}{\mathrm{\partial <!--\; \partial \; -->}{x}^2}(2)$
Again, $\frac{\mathrm{\partial <!--\; \partial \; -->}\mathrm{\Psi <!--\; \Psi \; -->}}{\mathrm{\partial <!--\; \partial \; -->}t}=-<!--\; -\; -->i\frac{E}{\mathrm{\hslash <!--\; \hslash \; -->}}\mathrm{\Psi <!--\; \Psi \; -->}⟹<!--\; ⟹\; -->E\mathrm{\Psi <!--\; \Psi \; -->}=-<!--\; -\; -->i\mathrm{\hslash <!--\; \hslash \; -->}\frac{\mathrm{\partial <!--\; \partial \; -->}\mathrm{\Psi <!--\; \Psi \; -->}}{\mathrm{\partial <!--\; \partial \; -->}t}$
Here we see operator $E=-<!--\; -\; -->i\mathrm{\hslash <!--\; \hslash \; -->}\frac{\mathrm{\partial <!--\; \partial \; -->}}{\mathrm{\partial <!--\; \partial \; -->}t}⟹<!--\; ⟹\; -->E^2=-<!--\; -\; -->{\mathrm{\hslash <!--\; \hslash \; -->}}^2\frac{{\mathrm{\partial <!--\; \partial \; -->}}^2}{\mathrm{\partial <!--\; \partial \; -->}{t}^2}$
$⟹<!--\; ⟹\; -->E^2\mathrm{\Psi <!--\; \Psi \; -->}=-<!--\; -\; -->{\mathrm{\hslash <!--\; \hslash \; -->}}^2\frac{{\mathrm{\partial <!--\; \partial \; -->}}^2\mathrm{\Psi <!--\; \Psi \; -->}}{\mathrm{\partial <!--\; \partial \; -->}{t}^2}(3)$
Combining (2) and (3) we find the differential equation:
$\frac{{\mathrm{\partial <!--\; \partial \; -->}}^2\mathrm{\Psi <!--\; \Psi \; -->}}{\mathrm{\partial <!--\; \partial \; -->}{x}^2}=\frac{1}{C^2}\frac{{\mathrm{\partial <!--\; \partial \; -->}}^2\mathrm{\Psi <!--\; \Psi \; -->}}{\mathrm{\partial <!--\; \partial \; -->}{t}^2}$
It is the Maxwell's equation, not the well known Schrodiner equation!
Therefore for the Schrodinger equation to exist, the de Broglie approximation must hold for $m_0\ne <!--\; \ne \; -->0.$ I see a clear contradiction here. Then why is the Schrodinger equation correct after all?

Why doesn't calculating the de Broglie wavelength work with h in eV·s?
I'm trying to calculate the de Broglie wavelength of a particle with a known momentum $p=980.93\mathrm{G}\mathrm{e}\mathrm{V}/\mathrm{s}$. The de Broglie relation is:
$\mathrm{\lambda <!--\; \lambda \; -->}=\frac{h}{p}$
but I notice that if I use the value of $4.136\times <!--\; \times \; -->{10}^{-<!--\; -\; -->15}\mathrm{e}\mathrm{V}\cdot <!--\; \cdot \; -->\mathrm{s}$ for the Planck's constant, the answer that I get from the equation is in units of ${\mathrm{s}}^2$, not m. Specifically:
$\mathrm{\lambda <!--\; \lambda \; -->}=\frac{4.136\times <!--\; \times \; -->{10}^{-<!--\; -\; -->15}\mathrm{e}\mathrm{V}\cdot <!--\; \cdot \; -->\mathrm{s}}{980.93\mathrm{G}\mathrm{e}\mathrm{V}/\mathrm{s}}=4.25\times <!--\; \times \; -->{10}^{-<!--\; -\; -->27}{\mathrm{s}}^2$
which is not at all equivalent to the value that one gets when using $h=6.626\times <!--\; \times \; -->{10}^{-<!--\; -\; -->34}\mathrm{J}\cdot <!--\; \cdot \; -->\mathrm{s}$ instead (since $\mathrm{J}\cdot <!--\; \cdot \; -->\mathrm{s}$ can be converted into $\frac{\mathrm{k}\mathrm{g}\cdot <!--\; \cdot \; -->{\mathrm{m}}^2}{{\mathrm{s}}^2}\cdot <!--\; \cdot \; -->\mathrm{s}$).
$\mathrm{\lambda <!--\; \lambda \; -->}=\frac{6.626\times <!--\; \times \; -->{10}^{-<!--\; -\; -->34}\frac{\mathrm{k}\mathrm{g}\cdot <!--\; \cdot \; -->{\mathrm{m}}^2}{{\mathrm{s}}^2}\cdot <!--\; \cdot \; -->\mathrm{s}}{5.237\times <!--\; \times \; -->{10}^{-<!--\; -\; -->16}\frac{\mathrm{k}\mathrm{g}\cdot <!--\; \cdot \; -->\mathrm{m}}{\mathrm{s}}}=1.27\times <!--\; \times \; -->{10}^{-<!--\; -\; -->18}\mathrm{m}$
What's going on here?

How do you change Planck's law from frequency to wavelength?
I have to derive Wien's displacement law by using Planck's law. I've tried but I come to a unsolvable equation (well I can't solve it) anywhere I look online it comes to the same conclusion, you need to solve an equation involving transcendentals which I have no idea what they are nor any of the math classes required for this physics course teach it. Once that equation is solved, the next steps are quite simple, by using the solution.
Maybe my professor just wants us to google the solution to the equation and used it to keep solving the problem?
anyway, what really bothers me is this:
My book shows Planck's law in terms of frequency
$B_{\mathrm{\nu <!--\; \nu \; -->}}(T)=\frac{2h{\mathrm{\nu <!--\; \nu \; -->}}^3}{c^2}\frac{1}{e^{h\mathrm{\nu <!--\; \nu \; -->}/(k_{B}T)}-<!--\; -\; -->1}$
but to do the problem I need it in terms of wavelength
$B_{\mathrm{\lambda <!--\; \lambda \; -->}}(T)=\frac{2h{c}^2}{{\mathrm{\lambda <!--\; \lambda \; -->}}^5}\frac{1}{e^{hc/(\mathrm{\lambda <!--\; \lambda \; -->}{k}_{B}T)}-<!--\; -\; -->1}$
And $v=c/\mathrm{\lambda <!--\; \lambda \; -->}$, so by that then it can be said that the exponent of $\mathrm{\lambda <!--\; \lambda \; -->}$ would be 3 when it is expressed in terms of lambda (plus $c$ would have an exponent of 1). But Wikipedia says it is 5 and not 3. It multiplies by the derivative of $\mathrm{\nu <!--\; \nu \; -->}$, but I don't get why. I'm just substituting, not differentiating, no need to use the chain rule.

Why is a relativistic calculation needed unless $pc$ is much smaller than the rest energy of a particle?
After introducing the de Broglie wavelength equation, my textbook gives a rather simple example where it asks to find the kinetic energy of a proton whose de Broglie wavelength is 1 fm. In the solution to this problem, it states that "A relativistic calculation is needed unless $pc$ for the proton is much smaller than the proton rest energy."
Could someone please explain why this is the necessary condition? I'm not sure what the quantity '$pc$' represents or means here. I know that for massless particles like photons, the total energy $E$ is equal to $pc$. I'm not sure what it means for particles having rest mass like protons.

What causes hot things to glow, and at what temperature?
I have an electric stove, and when I turn it on and turn off the lights, I notice the stove glowing.
However, as I turn down the temperature, it eventually goes away completely. Is there a cut-off point for glowing?
What actually is giving off the light? Does the heat itself give off the light, or the metal?

Relativistic change of mass and De-broglie's equation
From De Broglie's equation
$\mathrm{\lambda <!--\; \lambda \; -->}=\frac{h}{mv}$
We can see that $\mathrm{\lambda <!--\; \lambda \; -->}$ depends inversely on $m$
My question is:
If the $m$ in the equation indicates the kinetic mass of the body then from this relation
$m=\frac{m_0}{\surd (1-<!--\; -\; -->v^2/c^2)}$
$m$ would increase with $v$. Thus the $\mathrm{\lambda <!--\; \lambda \; -->}$ value should change doubly...first for the rise of $v$ and second for the rise of $m$
Or is the $m$ in De Broglie's equation representing rest mass, meaning that I am altogether wrong.
I am a layman in Physics and I stumbled on this question while understanding electron microscopy

Calculating an energy of an electron with known De Broglie wavelength (why can't we calculate it similar than we do it for a photon)
Lets say we have an electron with known De Broglie wavelength $\mathrm{\lambda <!--\; \lambda \; -->}$. Can anyone justify or explain why we calculate its energy $E$ using 1st the De Broglie relation $\mathrm{\lambda <!--\; \lambda \; -->}=h/p$ to get momentum $p$ and 2nd using the invariant interval to calculate $E$
$\begin{array}{rl}{p}^2{c}^2& =E^2-<!--\; -\; -->{E_0}^2\\ E& =\sqrt{p^2{c}^2+{E_0}^2}\end{array}$
Why we are not alowed to do it like we do it for a photon:
$\begin{array}{r}E=h\mathrm{\nu <!--\; \nu \; -->}=h\frac{c}{\mathrm{\lambda <!--\; \lambda \; -->}}\end{array}$
These equations return different results.

Relationship between Wien's law and Stefan-Boltzmann's law
I'm studying Quantum mechanics by Bransden and Joachain and in the introduction chapter it says:
Wien showed that the spectral distribution function had to be on the form
$\mathrm{\rho <!--\; \rho \; -->}(\mathrm{\lambda <!--\; \lambda \; -->},T)={\mathrm{\lambda <!--\; \lambda \; -->}}^{-<!--\; -\; -->5}f(\mathrm{\lambda <!--\; \lambda \; -->}T)$
where $f(\mathrm{\lambda <!--\; \lambda \; -->}T)$ is a function of the single variable $\mathrm{\lambda <!--\; \lambda \; -->}T$. It is a simple matter to show that Wien's law includes Stefan-Boltzmann law $R(T)=\mathrm{\sigma <!--\; \sigma \; -->}{T}^4$
One of the exercises is to show this and I cannot understand how to.
This is what I've tried:
The relationship between spectral emittance and spectral distribution is
$\mathrm{\rho <!--\; \rho \; -->}(\mathrm{\lambda <!--\; \lambda \; -->},T)=\frac{4}{c}R(\mathrm{\lambda <!--\; \lambda \; -->},T),$
where c is the speed of light, which inserted in the above equation gives
$R(\mathrm{\lambda <!--\; \lambda \; -->},T)=\frac{c}{4}{\mathrm{\lambda <!--\; \lambda \; -->}}^{-<!--\; -\; -->5}f(\mathrm{\lambda <!--\; \lambda \; -->}T).$
Now, the total spectral emittance is the integral of $R$ over all wavelengths so
$R(T)=\frac{c}{4}\underset{0}{\overset{\mathrm{\infty <!--\; \infty \; -->}}{\int <!--\; \int \; -->}}{\mathrm{\lambda <!--\; \lambda \; -->}}^{-<!--\; -\; -->5}f(\mathrm{\lambda <!--\; \lambda \; -->}T)d\mathrm{\lambda <!--\; \lambda \; -->}$
This is where I'm stuck. Can anyone help me figure this out?

Where am I doing mistake of concept in deriving the De Broglie equation?
When I am deriving de Broglie wavelength for a relativistic particle using $E^2=m^2{c}^4+p^2{c}^2$ and equating with $E=\frac{hc}{\mathrm{\lambda <!--\; \lambda \; -->}}$, and then putting $p=kmv$, $k$ being relativistic factor, I am getting $\mathrm{\lambda <!--\; \lambda \; -->}=\frac{h}{kmc}$ instead of $\frac{h}{kmv}$.
Is there any mistake that i am doing with equating those 2 energy equations ? Or something else ?